Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $z = \dfrac{2n + 10}{n^2 - 2n} \times \dfrac{n^3 - 8n^2 + 12n}{-2n^2 - 14n - 20} $
Answer: First factor out any common factors. $z = \dfrac{2(n + 5)}{n(n - 2)} \times \dfrac{n(n^2 - 8n + 12)}{-2(n^2 + 7n + 10)} $ Then factor the quadratic expressions. $z = \dfrac {2(n + 5)} {n(n - 2)} \times \dfrac {n(n - 2)(n - 6)} {-2(n + 5)(n + 2)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {2(n + 5) \times n(n - 2)(n - 6) } {n(n - 2) \times -2(n + 5)(n + 2) } $ $z = \dfrac {2n(n - 2)(n - 6)(n + 5)} {-2n(n + 5)(n + 2)(n - 2)} $ Notice that $(n + 5)$ and $(n - 2)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {2n(n - 2)(n - 6)\cancel{(n + 5)}} {-2n\cancel{(n + 5)}(n + 2)(n - 2)} $ We are dividing by $n + 5$ , so $n + 5 \neq 0$ Therefore, $n \neq -5$ $z = \dfrac {2n\cancel{(n - 2)}(n - 6)\cancel{(n + 5)}} {-2n\cancel{(n + 5)}(n + 2)\cancel{(n - 2)}} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $z = \dfrac {2n(n - 6)} {-2n(n + 2)} $ $ z = \dfrac{-(n - 6)}{n + 2}; n \neq -5; n \neq 2 $